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dybincka [34]
2 years ago
13

Why is wood a renewable resource?

Physics
1 answer:
sweet-ann [11.9K]2 years ago
6 0

Answer:

Trees grow everywhere, so we always have some

Explanation:

Wood is a renewable resource; trees can be replanted and grown to maturity in place of those that are cut down. As long as the trees are replanted at the same rate as they are cut down wood will be a renewable resource.

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Sort the statements about heat transfer into the correct columns.
Hoochie [10]

Answer: Conduction- Touch transfer heat and Earth warms air

Convection- liquid/gas transfers heat and warm air rises

Radiation- Sun heats Earth and Waves transfer heat

Explanation:

5 0
3 years ago
Read 2 more answers
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0
3 years ago
If an electron is accelerated from rest through a potential difference of 1200V find its approximate velocity at the end of this
kolbaska11 [484]

Answer: 2.1 × 10^7 m/s

Explanation:

Please see the attachments below

8 0
3 years ago
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